Question

Solve the following equations by method of inversion:

x + y + z = 1, x – y + z = 2 and x + y – z = 3

Answer 1

Matrix form of the given system of equations is

`[(1,1,1),(1, -1, 1),(1, 1, -1)][(x),(y),(z)] = [(1),(2),(3)]`

This is of the form AX = B,

where A = `[(1,1,1), (1,-1,1),(1, 1, -1)], "X" = [(x),(y),(z)] "and B"= [(1),(2),(3)]`

Let us find A⁻¹.

|A| = `[(1,1,1),(1,-1,1),(1,1, -1)]`

= 1(1 − 1) −1(−1 − 1) + 1(1 + 1)

= 0 + 2 + 2

= 4 ≠ 0

∴ A⁻¹ exists.

Consider AA^–1 = I

∴ `[(1,1,1),(1, -1, 1),(1, 1, -1)]"A"^-1 = [(1, 0, 0),(0, 1, 0),(0, 0, 1)]`

Appying R₂ – R₁ and R₃ – R₁, we get

`[(1, 1, 1),(0, -2, 0),(0, 0, -2)]"A"^-1 = [(1, 0, 0),(-1, 1, 0),(-1, 0, 1)]`

Appying `(-1/2)` R₂, and `(-1/2)` R₃, we get

`[(1, 1, 1),(0, 1, 0),(0, 0, 1)]"A"^-1 = [(1, 0, 0),(1/2, -1/2, 0),(1/2, 0, -1/2)]`

Appying R₁ – R_2, we get

`[(1, 0, 1),(0, 1, 0),(0, 0, 1)]"A"^-1 = [(1/2, 1/2, 0),(1/2, -1/2, 0),(1/2, 0, -1/2)]`

Appying R₁ – R_3, we get

`[(1, 0, 0),(0, 1, 0),(0, 0, 1)]"A"^-1 = [(0, 1/2, 1/2),(1/2, -1/2, 0),(1/2, 0, -1/2)]`

∴ A^–1  = `1/2[(0, 1, 1),(1, -1, 0),(1, 0, -1)]`

pre-multiplying AX = B by A^–1, we get

A^–1(AX) = A^–1B

∴ (A^–1A)X = A^–1B

∴ IX = A^–1B

∴ X = `1/2[(0,1,1),(1,-1,0),(1,0,-1)][(1),(2),(3)]`

∴ `[(x),(y),(z)] = 1/2[(0 + 2 + 3),(1 -2 + 0),(1 +0 -3)]`

= `1/2[(5),(-1),(-2)]`

∴ `[(x),(y),(z)] = [(5/2),(-1/2),(-1)]`

By equality of matrices, `x = 5/2`, `y = -1/2`, z = −1 is the required solution.